what is the equation describing the motion of mass on the end of a spring which is streached 8.8 cm from equlibrium and then released from rest and whose period is .66 s ? what will be its displacement after 1.8 s
To determine the equation describing the motion of a mass on the end of a spring, we can use Hooke's Law and the equation for simple harmonic motion.
1. Hooke's Law states that the force exerted by a spring is directly proportional to the displacement from equilibrium: F = -kx, where F is the force, k is the spring constant, and x is the displacement from equilibrium.
2. The equation for simple harmonic motion is: x(t) = A*cos(2πft + φ), where x(t) is the displacement at time t, A is the amplitude (maximum displacement), f is the frequency (1/period), and φ is the phase constant.
Now, let's find the values for these parameters step by step:
1. We are given that the spring is stretched 8.8 cm from equilibrium. The displacement (A) is the amplitude, so A = 0.088 m.
2. We are also given that the period is 0.66 s. The frequency (f) is the reciprocal of the period, so f = 1/0.66 ≈ 1.515 Hz.
3. The phase constant (φ) represents the initial phase of the motion. Since the mass is released from rest, it starts at equilibrium. This means φ = 0.
Now, we can plug in these values into the equation for simple harmonic motion:
x(t) = 0.088*cos(2π*1.515t)
To find the displacement after 1.8 s, we substitute t = 1.8 into the equation:
x(1.8) = 0.088*cos(2π*1.515*1.8)
Using a calculator, evaluate the expression to find the displacement after 1.8 s.